A 1.0 M solution of `Cd^(2+)` is added to excess iron and the system is allowed to reach equillibrium. What is the concentration of `Cd^(2+)` ?
`Cd^(2+)(aq)+Fe(s)toCd(s)+Fe^(2+)(aq), E^(@)=0.037`

A 1.0 M solution of Cd^(+2) is added to excess iron and the system is allowed to reach equilibrium . What is the concentration of Cd^(+2) ? Cd^(2+)(aq) + Fe(s) to Cd(s) + Fe^(2+) (aq) , E^(0) = 0.037

Write cell representation for following cells Cd^(2+)(aq)+Zn(s)toZn^(2+)(aq)+Cd(s)

Write the Nernst equation for the reaction : 2Cr(s)+3Cd^(2+)(aq) to 2Cr^(3+)(aq)+3Cd(s)

Calculate the equilibrium constant for the reaction 2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2(s) E0cell = 0.236V

Write Nernst equation for the reaction: (i) 2Cr(s) +3Cd^(2+)(aq) to 2Cr^(3+) (aq) +3Cd(s)

Calculate the equilibrium constant for the reaction Fe(s)+Cd2+(aq)⇔Fe2+(aq)+Cd(s) (Given : E∘Cd2+∣Cd=0.40V,E∘Fe2+∣Fe=−0.44V).

(a) Express the relationship amongst cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solute related to conductivity of its solution ? (b) Calculate the equilibrium constant for the reaction Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd (s) (Given : E_(Cd^(2+)|Cd)^(@) = 0.40 V , E_(Fe^(2+) |Fe)^(@) = -0.44 V ).

Calculate the equilibrium constant for the reaction: Cd^(2+)+Zn(s)rarrZn^(2+)(aq)+Cd(s) If E_(Cd^(2+)|Cd)=-0.403V and E_(Zn^(2+)|Zn)=-.763V

Calculate the equilibrium constant for the reaction : Fe(s)Cd^(2+)(aq)harrFe^(2+)(aq)+Cd(s) ("Given" E_(cd^(2+)//Cd)^(@)=-0.40 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V)